An organic compound [X] reacts with \(H_2/Pd-BaSO_4\) to give compound [Y], which reduces Tollens' reagent and undergoes Cannizzaro's reaction. On rigorous oxidation of [Y] in presence of \(KMnO_4/H^+\), phthalic acid is the product obtained. What is [X]?
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Rosenmund reduction converts acid chlorides into aldehydes. Aromatic aldehydes reduce Tollens' reagent and undergo Cannizzaro's reaction because they do not contain alpha hydrogen.
Step 1: Identify the reagent \(H_2/Pd-BaSO_4\).
The reagent \(H_2/Pd-BaSO_4\) is used in Rosenmund reduction.
In Rosenmund reduction, an acid chloride is selectively reduced to an aldehyde.
\[
RCOCl \xrightarrow{H_2/Pd-BaSO_4} RCHO
\]
So, compound [X] must be an acid chloride. Step 2: Check which option contains acid chloride group.
Among the given options, option (B) contains the group \(COCl\), which is an acid chloride group.
\[
\text{Option (B)} = o\text{-methylbenzoyl chloride}
\]
Therefore, option (B) is a strong possible answer. Step 3: Product formed by Rosenmund reduction.
When \(o\)-methylbenzoyl chloride undergoes Rosenmund reduction, it gives \(o\)-methylbenzaldehyde.
\[
o\text{-}CH_3C_6H_4COCl \xrightarrow{H_2/Pd-BaSO_4} o\text{-}CH_3C_6H_4CHO
\]
Thus, compound [Y] is \(o\)-methylbenzaldehyde. Step 4: Verify Tollens' test.
Aldehydes reduce Tollens' reagent and give silver mirror.
Since [Y] is an aromatic aldehyde, it reduces Tollens' reagent.
\[
o\text{-}CH_3C_6H_4CHO \xrightarrow{\text{Tollens' reagent}} \text{Silver mirror}
\]
So, this condition is satisfied. Step 5: Verify Cannizzaro's reaction.
Cannizzaro's reaction is shown by aldehydes which do not have alpha hydrogen.
Aromatic aldehydes such as benzaldehyde and substituted benzaldehydes do not have alpha hydrogen.
Therefore, \(o\)-methylbenzaldehyde undergoes Cannizzaro's reaction.
\[
2ArCHO + OH^- \rightarrow ArCOO^- + ArCH_2OH
\]
So, this condition is also satisfied. Step 6: Oxidation with \(KMnO_4/H^+\).
Strong oxidising agent \(KMnO_4/H^+\) oxidises both aldehyde group and benzylic methyl group into carboxylic acid groups.
In \(o\)-methylbenzaldehyde, the groups are \(CHO\) and \(CH_3\) at ortho positions.
Both groups get oxidised to \(COOH\):
\[
o\text{-}CH_3C_6H_4CHO \xrightarrow{KMnO_4/H^+} o\text{-}HOOC-C_6H_4-COOH
\]
The product is phthalic acid. Step 7: Final conclusion.
Since compound [X] must be an acid chloride which gives \(o\)-methylbenzaldehyde on Rosenmund reduction, and this aldehyde gives phthalic acid on oxidation, compound [X] is:
\[
\boxed{o\text{-methylbenzoyl chloride}}
\]
Hence, the correct answer is option (B).
